The integration of these functions can be performed by the substitution of sin 2 α {\displaystyle \sin ^{2}\alpha } or cos 2 α {\displaystyle \cos ^{2}\alpha } by a function in terms of cos 2 α {\displaystyle \cos 2\alpha } respectively.
By the double-angle formula, cos 2 α = 2 cos 2 α − 1 {\displaystyle \left.\cos 2\alpha =2\cos ^{2}\alpha -1\right.} . Rearranging to make cos 2 α {\displaystyle \cos ^{2}\alpha } the subject, cos 2 α = 1 2 + 1 2 cos 2 α {\displaystyle \cos ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha } .
We know by the Pythagorean identity sin 2 θ + cos 2 θ = 1 {\displaystyle \left.\sin ^{2}\theta +\cos ^{2}\theta =1\right.} that cos 2 α = 1 − sin 2 α {\displaystyle \left.\cos ^{2}\alpha =1-\sin ^{2}\alpha \right.} , so 1 − sin 2 α = 1 2 + 1 2 cos 2 α {\displaystyle 1-\sin ^{2}\alpha ={\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha } . Rearranging, sin 2 α = 1 2 − 1 2 cos 2 α {\displaystyle \sin ^{2}\alpha ={\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha } .
By substituting sin 2 α {\displaystyle \sin ^{2}\alpha \,} for 1 2 − 1 2 cos 2 α {\displaystyle {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha } , ∫ sin 2 α d α = ∫ 1 2 − 1 2 cos 2 α d α {\displaystyle \int \sin ^{2}\alpha d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha d\alpha } . Similarly for cos 2 α {\displaystyle \cos ^{2}\alpha \,} ,
∫ cos 2 α d α = ∫ 1 2 + 1 2 cos 2 α d α {\displaystyle \int \cos ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha }
These can be integrated by using the primitives of cosine and sine
∫ cos α d α = sin α + c {\displaystyle \int \cos \alpha d\alpha =\sin \alpha +c}
∫ sin α d α = − cos α + c {\displaystyle \int \sin \alpha d\alpha =-\cos \alpha +c}
For cos2 α
∫ 1 2 + 1 2 cos 2 α d α = 1 2 α + 1 4 sin 2 α + c {\displaystyle \int {\tfrac {1}{2}}+{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c}
∫ cos 2 α d α = 1 2 α + 1 4 sin 2 α + c {\displaystyle \int \cos ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha +{\tfrac {1}{4}}\sin 2\alpha +c}
For sin2 α
∫ sin 2 α d α = ∫ 1 2 − 1 2 cos 2 α d α {\displaystyle \int \sin ^{2}\alpha \,d\alpha =\int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha }
∫ 1 2 − 1 2 cos 2 α d α = 1 2 α − 1 4 sin 2 α + c {\displaystyle \int {\tfrac {1}{2}}-{\tfrac {1}{2}}\cos 2\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c}
∫ sin 2 α d α = 1 2 α − 1 4 sin 2 α + c {\displaystyle \int \sin ^{2}\alpha \,d\alpha ={\tfrac {1}{2}}\alpha -{\tfrac {1}{4}}\sin 2\alpha +c}
for 
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. Similarly for 
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